Because it is linear (x + 5), quick observation tells us it is increasing along the interval 4 < x < 6
meaning no y-value will ever be greater than 11 along this interval for this equation, hence

0 < |x - 5| < δ → |x - 5||x + 5| < |x - 5| 11 < ε

and the proof can be shown

My question is: Lets say you solve the quadratic and get imaginary roots, imaginary equations. Lets say that h(x) is a function with imaginaries within. How would I bound that function by a δ of 1, or by anything for that matter, to make sure it is not increasing without bounds?

You can always defined an order (<,S)on any set S ; a partial order is a relationship on the set that is
i)Reflexive
ii)Antisymmetric
iii) Transitive

If any two elements z,w are comparable, i.e., if you can always decide whether z<w or w<z , then the order is
a total order; otherwise it is a partial order. Examples of partial order, maybe the most natural one, is that
defined on the collection of subsets by inclusion. Maybe the best example of a total order is that of the Reals
with a>b => a-b >0 .

Now, the Reals under the standard < , i.e., a<b iff (Def.) b-a>0 are an ordered field. Notice the problem comes
from the imaginary numbers (assuming the Real part) : if i>0 , then i.i =-1 . Now you can then assume in
your theory that -1>0 . But then (-1)(-1)=1 >0 , so 1>0 and -1>0 . But then 1+(-1)=0 >0

Check out the linked page to see how it is not possible to turn the usual Complex numbers; usual sum and product, into an ordered field.

Basically , a<b can mean many different things depending on the context.